`0.1` mol each of ethyl alcohol and acetic acid are allowed to react and at equilibrium, the acid was exactly neutralised by `100 mL` of `0.85 N NaOH`. If no hydrolysis of ester is supposed to have undergo, finf `K_(c)`.

Answer

C

Explanation

`{:(CH_(3)COOH,+,C_(2)H_(5)OH,hArr,CH_(3)COOC_(2)H_(5),+,H_(2)O),(0.1,,0.1,,0,,0),(,,,,,,"Before reaction"),((0.1-x),,(0.1-x),,x,,x):}`
mEq of acetic aid left = mEq of NaOH used
`=100xx0.85=85`
`:.` Milli moles of aetic acid left `=85` ( `:'` mono besic)
`:.` moles of acetic acid left`=0.085`
or `(0.01-x)=0.085`
`x=0.015`
Now, `K_(C)=x^(2)/((0.1-x)^(2))=((0.015)^(2))/((0.085)^(2))=0.031`